It seems the following.

In general the answer is no, because compactness does not imply sequential compactness. Let $\Bbb T=\{z\in\Bbb C:|z|=1\}$ be the unit circle endowed with the standard topology. Put $G={\Bbb T}^{\Bbb T}$. By Tychonov Theorem, $G$ is a compact space. Let $K=\{e\}$ be the trivial subgroup of $G$. Select an element $g=(g_z)_{z\in\Bbb T}\in G$ such that $g_z=z$ for each $z\in\Bbb T$. Suppose that there exists an increasing sequence $\{i_n\}$ of positive integers such that the sequence $\{g^{i_n}\}$ converges to the unit of the group $G$. Let $U_0=\{z\in\Bbb T: \operatorname{Re} z\ge 0\}$ be a neighborhood of the unit of the group $\Bbb T$. For each angle $z\in \Bbb T $ we can choose a number $n_z$ such that $i_nz\in U_0$ for each $n>n_z$. For each natural number $n$ put $G_n=\{z\in\Bbb T:n_z=n\}$. The continuity of power on the group $\Bbb T$ implies that the set $G_n$ is closed for each natural number $n$. Since $G=\bigcup_{n\in\Bbb N} G_n$, Baire Theorem implies that there exists a number $m$ such that a set $G_m$ has non-empty interior. Therefore there exists an open arc $U\subset G_m$ of the circle $\Bbb T$. Since the sequence $\{i_n\}$ is increasing, there exists a number $n>m$ such that $i_n>1/\mu(U)$, where $\mu$ is the standard measure on $\Bbb T$ such that $\mu(\Bbb T)=1$. But then $U_0\supset i_nG_m\supset i_n\overline U=\Bbb T$, a contradiction.

I hope you will be almost happy with this **addendum**. Let $G$ be a Hausdorff totally disconnected and locally compact topological group, $K$ be a cocompact *normal* subgroup of the group $G$ and $g\in G$. For each $n$ put $i_n=n!$. I claim that a sequence $\{g^{i_n}K\}$ converges to $K$ in the coset space $G/K$. Indeed, by [Pon, Theorem 16], the group $G$ has a base $\mathcal B$ at the unit consisting of its open compact subgroups. Let $H\in\mathcal B$ be an arbitrary group. Since the group $K$ is normal then $HK$ is a group. Since $\{hHK: h\in G\}$ is an open cover of the compact space $G/K$, there exists a finite subset $F$ of the group $G$ such that $G=\bigcup\{hHK: h\in G\}$. Then the pigeonhole principle implies that there exist natural numbers $k<l$ and an element $h\in F$ such that $g^k,g^l\in hHK$. Then $g^{l-k}\in K^{-1}H^{-1}h^{-1}hHK=KHK=HK$. Since the set $HK$ is a group, $g^{i_n}\in HK$ for each $n\ge l-k$.

[Pon] Lev S. Pontrjagin, *Continuous groups*, 2nd ed., M., (1954) (in Russian).